A group of 9 students, is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that cannot be selected for the team X, and cannot be selected for the team Y. Then the number of ways to form such teams is ________.

Question

A group of 9 students, $s_{1}, s_{2}, \dots, s_{9}$ is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that $s_{1}$ cannot be selected for the team X, and $s_{2}$ cannot be selected for the team Y. Then the number of ways to form such teams is ________. [2024]

Smart Achievers · Accepted Answer

(665)

Number of required ways

$= \frac{9!}{2! 3! 4!} - (n (s_{1} \in X) + n (s_{2} \in Y) - n (s_{1} \in X and s_{2} \in Y))$

$= \frac{9!}{2! 3! 4!} - (\frac{8!}{1! 3! 4!} + \frac{8!}{2! 2! 4!} - \frac{7!}{1! 2! 4!}) = 665$

Solution

Ans.
(665)

Number of required ways

$= \frac{9!}{2! 3! 4!} - (n (s_{1} \in X) + n (s_{2} \in Y) - n (s_{1} \in X and s_{2} \in Y))$

$= \frac{9!}{2! 3! 4!} - (\frac{8!}{1! 3! 4!} + \frac{8!}{2! 2! 4!} - \frac{7!}{1! 2! 4!}) = 665$

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Solution

Q. A group of 9 students, s1,s2,…,s9 is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that s1 cannot be selected for the team X, and s2 cannot be selected for the team Y. Then the number of ways to form such teams is ________. [2024]

Ans. (665) Number of required ways =9!2!3!4!-(n(s1∈X)+n(s2∈Y)-n(s1∈X and s2∈Y)) =9!2!3!4!-(8!1!3!4!+8!2!2!4!-7!1!2!4!)=665

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Ans.
(665)

Number of required ways

$= \frac{9!}{2! 3! 4!} - (n (s_{1} \in X) + n (s_{2} \in Y) - n (s_{1} \in X and s_{2} \in Y))$

$= \frac{9!}{2! 3! 4!} - (\frac{8!}{1! 3! 4!} + \frac{8!}{2! 2! 4!} - \frac{7!}{1! 2! 4!}) = 665$